$x=4 \cos \left(8 t-\frac{y}{2}\right)$
Comparing with $x=A \sin (k x-\omega t)$
$\omega=8$
and $\omega=2 \pi f$
$f=\frac{8}{2 \pi}$
$f=\frac{4}{\pi}$

(Given, $\mathrm{R}=8.3 \mathrm{JK}^{-1}, \gamma=1.4$ )
$(A)$ $u=0.8 v$ and $f_5=f_0$
$(B)$ $u=0.8 v$ and $f_5=2 f_0$
$(C)$ $u=0.8 v$ and $f_5=0.5 f_0$
$(D)$ $u=0.5 v$ and $f_5=1.5 f_0$